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Explain what would happen if in the capacitor given in 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates C=1.771 x 10^-11 F, after the supply voltage of 100 V is was disconnected

$\begin{array}{1 1} 10 V \\ 100\;V \\ 16.7\;V \\ 50\;V\end{array} $

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A)
Solution :
Dielectric constant, k = 6
Initial capacitance, $C = 1.771 × 10^{−11} F$
New capacitance, $C_1 = kC = 6 × 1.771 × 10^{−11} F = 106 pF$
If supply voltage is removed, then there will be constant amount of charge in the plates.
Charge $= 1.771 × 10^{−9} C $
Potential across the plates is given by
$V_1= \large\frac{q}{C_1} =\frac{1.771 \times 10^{-9}}{106 \times 10^{12} }$$=16.7\;V$
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